已知数列{an}的前n项和为Sn,且Sn=1/3(an−1),求证数列{an}为等比数列,并求其通项公式.

问题描述:

已知数列{an}的前n项和为Sn,且Sn

1
3
(an−1),求证数列{an}为等比数列,并求其通项公式.

Sn

1
3
(an−1)可知Sn−1
1
3
(an−1−1)

两式相减可得,an
1
3
(anan−1)

an
an−1
=−
1
2
,(n≥2)
故数列数列{an}为等比数列.公比q=
1
2

 又a1S1
1
3
(a1−1)

a1=−
1
2

an=(−
1
2
)n